Ta có:
\(\dfrac{1}{2}\left(cos4x+cos2x\right)-cos4x=2-4sin^2\left(\dfrac{\pi}{4}-\dfrac{3x}{2}\right)\)
<=> \(\dfrac{1}{2}cos4x+\dfrac{1}{2}cos2x-cos4x=2\left(1-2sin^2\left(\dfrac{n}{4}-\dfrac{3x}{2}\right)\right)\)
<=> \(\dfrac{1}{2}cos2x-\dfrac{1}{2}cos4x=2cos\left(\dfrac{n}{2}-3x\right)\)
<=> \(\dfrac{1}{2}\left(cos2x-cos4x\right)=2sin3x\)
<=> \(\dfrac{1}{2}\left(-2\right)sin3xsin\left(-x\right)=2sin3x\)
<=>\(\dfrac{1}{2}\left(-2\right)\left(-1\right)sin3xsinx=2sin3x\)
<=>\(sin3xsinx=2sin3x\)
<=> \(sin3xsinx-2sin3x=0\)
<=>\(sin3x\left(sinx-2\right)=0\)
<=> \(\left[{}\begin{matrix}sin3x=0\\sinx-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=k\pi,k\in Z\\sinx=2\end{matrix}\right.\)
Do sin\(\in\left[-1,1\right]\)nên sinx =2 ( loại)
\(3x=k\pi\Leftrightarrow x=k\dfrac{\pi}{3},k\in Z\)
\(cos3x.cosx-cos4x=2-4sin^2\left(\dfrac{\pi}{4}-\dfrac{3x}{2}\right)\)
\(\Leftrightarrow\dfrac{1}{2}cos4x+\dfrac{1}{2}cos2x-cos4x=2cos\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow\dfrac{1}{2}cos2x-\dfrac{1}{2}cos4x=2sin3x\)
\(\Leftrightarrow sin3x.sinx=2sin3x\)
\(\Leftrightarrow sin3x.\left(sinx-2\right)=0\)
\(\Leftrightarrow sin3x=0\)
\(\Leftrightarrow3x=k\pi\)
\(\Leftrightarrow x=\dfrac{k\pi}{3}\)
